Question

David is trying to solve the expression :
\[\frac{(4)^2 \times 2^{(n+1)} - 4 \times 2^n}{(4)^2 \times 2^{(n+2)} - 2 \times 2^{(n+2)}}\]
And you help him to do the same and finally arrive at the answer with correct to one decimal which would be - (Note:- DO NOT include spaces in your answer)

Answer 1

Correct Answer: 0.5

Step 1: Rewrite both parts using \( 2^{n+2} \).
Since \( 2^{n+1} = \frac{2^{n+2}}{2} \) and \( 2^n = \frac{2^{n+2}}{4} \), the numerator becomes \( 16 \times \frac{2^{n+2}}{2} - 4 \times \frac{2^{n+2}}{4} = 8 \times 2^{n+2} - 2^{n+2} = 7 \times 2^{n+2} \).

Step 2: Simplify the denominator the same way.
The denominator is \( 16 \times 2^{n+2} - 2 \times 2^{n+2} = 14 \times 2^{n+2} \).

Step 3: Divide.
\( \frac{7 \times 2^{n+2}}{14 \times 2^{n+2}} = \frac{7}{14} = 0.5 \). \[ \boxed{0.5} \]